/sci/ - Science & Math

[Anonymous No.5383512]

hi /sci/
i have assignment in algorithm analysis and design
to find the closed form solution for this

i got
3n^2 + 3n^3 + n
but i tried www.wolframalpha.com
and i got
n (1+3 n+3 n^3)
why is this

[Anonymous No.5383540]

links? steps?
get your shit together man

[Anonymous No.5383541]

so the left hand side is

<span class="math">=3n^3 + 2 \Sigma_{i=n+1}^{2n} i [/spoiler]

<span class="math">=3n^3 + 2 (\Sigma_{i=1}^{2n} i-\Sigma_{i=1}^{n} i) [/spoiler]

and

<span class="math"> \Sigma_{i=1}^{k} i=k(k+1)/2<span class="math">

Plug it in, everything works out as wolphram alpha says.[/spoiler][/spoiler]

[Anonymous No.5383542]

Well, the 3n^3 part doesn't depend on i, so that's immediately just n * 3n^3, and the 2i part is just 2 times the sum of n from 1 to 2n minus from 1 to n, so:
n*3n^3 + (2n*(2n+1)) - (n(n+1)) = n(3n^3 + 4n + 2 -n - 1) = n(3n^3 + 3n +1)
I am bad at explaining things.

[Anonymous No.5383543]

>>5383512
The answer you got appears to be the correct, except that you didn't multiply the 3n^3 by another n.

[Anonymous No.5383544]

>>5383543
no shit sherlock

[Anonymous No.5383545]

>>5383541
You made the same mistake he did. The sum of the 3n^3 terms is n * 3n^3, as it occurs that many times. Should be 3n^4

[Anonymous No.5383546]

The people helping OP are worse than OP

[Anonymous No.5383547]

>>5383545
true

[Anonymous No.5383548]

>>5383545
..true

[Anonymous No.5383574]

>>5383546

>No "religion vs science" threads

Yaaaaay, dogmatic behaviour from /sci/.